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Metric tensor with components

  1. Sep 7, 2014 #1


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    I have one question, which I don't know if I should post here again, but I found it in GR...
    When you have a metric tensor with components:
    [itex] g_{\mu \nu} = \eta _{\mu \nu} + h_{\mu \nu}, ~~ |h|<<1 [/itex] (weak field approximation).

    Then the inverse is:

    [itex] g^{\mu \nu} = \eta^{\mu \nu} - h^{\mu \nu} [/itex] right? However that doesn't give exactly that [itex] g^{\mu \rho}g_{\rho \nu} = \delta^{\mu}_{\nu} [/itex] because of the existence of the [itex]- h^{\mu \rho}h_{\rho \nu} [/itex] which is of course small but it's not zero... Can the inverse matrix be defined approximately?

    Also I don't understand why should the derivatives of [itex]h[/itex] behave as [itex]h[/itex] itself? I mean they take the terms like [itex] h \partial h, ~~ \partial h \partial h [/itex] to be of order [itex]\mathcal{O}(h^2)[/itex]... why?
    Last edited: Sep 7, 2014
  2. jcsd
  3. Sep 7, 2014 #2


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    They mean to write
    [tex]g_{ \mu \nu } = \eta_{ \mu \nu } + \epsilon h_{ \mu \nu } , \ \ \ |\epsilon | \ll 1.[/tex]
    Then all the following holds
    [tex]g \cdot g = \delta + \mathcal{O}( \epsilon^{2} ) , \ \ h \cdot h \sim h \partial h \sim \partial h \cdot \partial h \sim \mathcal{O} ( \epsilon^{ 2 } ) .[/tex]
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